In general if we have $$\sqrt{a_0+\sqrt{a_1+\sqrt{a_2....}}}$$ Are there easy ways of finding out if it converges, like from infinite products and series? Are there ways for converting the question of convergence of the radical to a one about series, like the logarithm does for products?
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5quick google gets: http://mathworld.wolfram.com/HerschfeldsConvergenceTheorem.html – muzzlator Mar 06 '13 at 16:26
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OK,thanks I've got it. – Ishan Banerjee Mar 06 '13 at 16:34
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According to the Herschfelds Convergence theorem, the sequence converges iff $a_n^{2^{-n}}$ converges, if $a_i>0 $. We can prove this by noting that, $$\sqrt{a_0+\sqrt{a_1+ \cdots + \sqrt{a_n}}}>{a_n^{2^{-n}}}$$ and also if $a_n<M^{2^{n}}$ then, $$\sqrt{a_0+\sqrt{a_1+ \cdots + \sqrt{a_n}}}<\sqrt{M}\sqrt{1+\sqrt{1 + \cdots}}$$ which is known to converge, and the sequence is non decreasing, showing that it converges. Much thanks to muzzlator for pointing out the theorem.

Zhanxiong
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Ishan Banerjee
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